A deterministic harvest scheduler using perfect bin-packing theorem

Abstract

The use of heuristic techniques in forest planning has been promoted by the need to solve complex problems that cannot be solved using mixed integer programming. We proved that for merchantability standards ensuring the perfect bin-packing theorem (PBPT), the maximum volume that can be harvested annually equals the sum of the maximum MAI of the stands. The method accommodates optimality criteria at the stand level, regarded as maximum MAI, and at the forest level, regarded as maximum annual allowable cut. We scheduled the harvesting by adjusting the first fit decreasing algorithm (FFD) to the PBPT conditions. When PBPT conditions were not met, we developed a mixed integer programming solution to adjust the merchantability standards of the stands to the distributional requirements of the PBPT, an adjustment that ensured the optimal performance of the FFD. The adjusted FFD was compared with linear programming (LP) and simulated annealing (SA) using two harvesting ages (i.e., one based on MAI maximization and one determined as the minimal age) and the same set of spatial temporal constraints for three areas in north-eastern British Columbia, Canada. We found that the adjusted FFD performed 100 times faster than SA and for annual allowable cut (AAC) supplied results that were more homogenous and at least 10% greater than the AAC supplied by SA. Furthermore, the adjusted FFD seemed to be relatively insensitive to spatial constraints (i.e., adjacency), while SA displayed a 70% reduction in AAC in response to an increase in adjacency delay from 1 year to 20 years. The results suggest that both adjusted FFD and SA are impacted by the selection of the harvesting age, but the adjusted FFD could still outperform SA.

Appendix

Corollary

For a predetermined set of merchantability standards ensuring PBPT fulfillment, the maximum even-flow volume that can be harvested annually is equal to the sum of the maximum mean annual increment of the stands.

Proof

The perfect packing theorem ensures that the MVHA can be reached by harvesting a selected yearly combination of stands if, and only if, the total volume of all stands at optimal harvesting age (OHA) is a multiple of MVHA. Therefore, the maximum amount of timber that can be harvested when perfect bin-packing conditions are fulfilled is

$$ \sum\limits_{j = 1}^{R} {\sum\limits_{i = 1}^{N} {X_{ij} V_{ij} = k \times {\text{MVHA}}} } $$

(5)

Where R is the rotation period, N is the number of stands used for MVHA computation, V _ij is the volume of stand i harvested at age j, X _ij is a binary variable identifying where or not the stand i was harvested at age j, and k is a positive integer ensuring the equality (5).

The planning period needs to be determined exactly as the bin-packing theorem operates in discrete non-stochastic settings. Therefore, concepts as the mean, median or mode of OHA cannot be used to represent forest rotation as some stands would require harvesting a non-integer number of times during rotation, violating PBPT settings. The perfect bin-packing theorem restricts R to values that ensure that all stands are harvested an integer number of times during rotation. The smallest possible number that satisfies this condition is the smallest common multiplier of the OHA. Thus, the forest rotation is determined as:

$$ R = R_{1}^{{\max \left( {a_{1}^{1} , \ldots ,a_{1}^{N} } \right)}} R_{2}^{{\max \left( {a_{2}^{1} , \ldots ,a_{2}^{N} } \right)}} \ldots R_{s}^{{\max \left( {a_{s}^{1} , \ldots ,a_{s}^{N} } \right)}} $$

(6)

where R ₁,…., R _s  = the decomposition factors of OHA for each stand, derived from numbers theory, a ⁱ _j  = the exponent of the decomposition factor R _j corresponding to OHA of stand i, OHA _i , ${\text{OHA}}_{i} = R_{1}^{{a_{1}^{i} }} R_{2}^{{a_{2}^{i} }} \ldots R_{s}^{{a_{s}^{i} }}$ for stand $i \in \{ 1,2 \ldots ,N\}$, s = the number of distinct decomposition factors among all OHA _i

The smallest common multiplier ensures that each stand is harvested an integer number of times during forest rotation but transforms R into a pseudo-rotation, as each stand will be harvested at least once in R years. The number of times stand i is harvested during this pseudo-rotation period is R/OHA _i :

$$ {\frac{R}{{{\text{OHA}}_{i} }}} = {\frac{{R_{1}^{{\max \left( {a_{1}^{1} , \ldots ,a_{1}^{N} } \right)}} R_{2}^{{\max \left( {a_{2}^{1} , \ldots ,a_{2}^{N} } \right)}} \ldots R_{s}^{{\max \left( {a_{s}^{1} , \ldots ,a_{s}^{N} } \right)}} }}{{R_{1}^{{a_{1}^{i} }} R_{2}^{{a_{2}^{i} }} \ldots R_{s}^{{a_{s}^{i} }} }}} = R_{1}^{{\max \left( {a_{1}^{1} , \ldots ,a_{1}^{N} } \right) - a_{1}^{i} }} R_{2}^{{\max \left( {a_{2}^{1} , \ldots ,a_{2}^{N} } \right) - a_{2}^{i} }} \ldots R_{s}^{{\max \left( {a_{s}^{1} , \ldots ,a_{s}^{N} } \right) - a_{s}^{i} }} \ge 1 $$

(7)

During the rotation period R the total harvested volume is $V = \sum\limits_{i = 1}^{N} {{\frac{R}{{{\text{OHA}}_{i} }}}} V_{{i,{\text{OHA}}_{i} }}$.

Therefore,

$$ {\text{MVHA}} = \frac{V}{R} = \left( {\sum\limits_{i = 1}^{N} {{\frac{R}{{{\text{OHA}}_{i} }}}} V_{{i,{\text{OHA}}_{i} }} } \right)/R = \sum\limits_{i = 1}^{N} {{\frac{{V_{{i,{\text{OHA}}_{i} }} }}{{{\text{OHA}}_{i} }}}} = \sum\limits_{i = 1}^{N} {{\text{MAI}}_{{i,{\text{OHA}}_{i} }} } = \sum\limits_{i = 1}^{N} {\max ({\text{MAI}}_{i} )} $$

(8)

where ${\text{MAI}}_{{i,{\text{OHA}}_{i} }}$ is the mean annual increment of stand i at OHA _i [q.e.d.]